Pumping Lemma

8. (10 pts) Use the pumping lemma to prove that language L is not regular. L = { 1nx | n > 0, x is a string over alphabet(0, 1)and x contains at most n Is } A string in L can be split into two parts: the first part contains only Is (let n denote the number of Is in the first part), and the second part z contains at most n 1s. For example, string 1111 e L because 1111 12x with x = 11 (or 11 11 = 13x with x = 1, or 11 11 = 14x with x = ). String 1101 L because 1101 = 12x with x = 01. String 1 11 101 E L because 111101 = 14x with x = 01 (or 111 101 = 13x with x = 101)